## Compute the semigroup of endomorphisms of an infinite cyclic semigroup

Compute $\mathsf{End}(S)$ where $S$ is an infinite cyclic semigroup.

We showed in this previous exercise that every infinite cyclic semigroup is isomorphic to $(\mathbb{N}^+, +)$. We now prove a lemma.

Lemma: Let $\varphi : S \rightarrow T$ be a semigroup homomorphism and suppose $A$ generates $S$. Then $\varphi$ is determined uniquely by its behavior on $A$ in the sense that if $\psi : S \rightarrow A$ is a semigroup homomorphism and $\psi(a) = \varphi(a)$ for all $a \in A$, then $\psi = \varphi$. Proof: Since $A$ generates $S$, every element of $S$ has the form $\prod a_i$ where $a : n \rightarrow A$ for some positive natural number $n$. Now $\varphi(\prod a_i) = \prod \varphi(a_i) = \prod \psi(a_i) = \psi(\prod a_i)$, so that $\psi = \varphi$. $\square$

In particular, $\mathbb{N}^+$ is generated (additively) by 1. So every endomorphism of $\mathbb{N}^+$ is determined uniquely by its image at 1.

On the other hand, if we define $\varphi_k$ on $\mathbb{N}^+$ by $\varphi_k(a) = ka$, we have $\varphi_k(a+b) = k(a+b)$ $= ka + kb$ $= \varphi_k(a) + \varphi_k(b)$. So $\varphi_k$ is an endomorphism of $\mathbb{N}^+$. Since $\varphi_k(1) = k$, by Lemma 1, if $\varphi$ is an endomorphism of $\mathbb{N}^+$ such that $\varphi(1) = k$, then $\varphi = \varphi_k$.

Define $\Psi : (\mathbb{N}^+, \cdot) \rightarrow \mathsf{End}(\mathbb{N}^+)$ by $\Psi(k) = \varphi_k$. Note that, for all $a \in \mathbb{N}^+$ (additive), we have $\Psi(k\ell)(a) = \varphi_{k\ell}(a)$ $= (k\ell) a$ $= k(\ell a)$ $= \varphi_k(\ell a)$ $= \varphi_k(\varphi_\ell(a))$ $= (\varphi_k \circ \varphi_\ell)(a)$ $= (\Psi(k) \circ \Psi(\ell))(a)$. So $\Psi(k\ell) = \Psi(k) \circ \Psi(\ell)$ for all $k,\ell \in \mathbb{N}^+$, and thus $\Psi$ is a semigroup homomorphism. Certainly $\Psi$ is surjective. Now if $\Psi(k) = \Psi(\ell)$, then in particular, $\Psi(k)(1) = \Psi(\ell)(1)$, so that $k = \ell$. Hence $\Psi$ is injective, and so an isomorphism.

Thus $\mathsf{End}((\mathbb{N}^+,+)) \cong (\mathbb{N}^+, \cdot)$.